Lagrange duality. The Lagrangian. We consider an optimization program of the form

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1 Lagrange duality Another way to arrive at the KKT conditions, and one which gives us some insight on solving constrained optimization problems, is through the Lagrange dual. The dual is a maximization program in λ, ν it is always concave (even when the original program is not convex), and gives us a systematic way to lower bound the optimal value. The Lagrangian We consider an optimization program of the form minimize f 0 (x) f m (x) 0, m = 1,..., M (1) x R N h p (x) = 0, p = 1,..., P. Much of what we will say below applies equally well to nonconvex programs as well as convex programs, so we will make it clear when we are taking the f m to be convex and the h p to be affine. We will take the domain of all of the f m and h p to be all of R N below; this just simplifies the exposition, we can easily replace this with the intersections of the dom f m and dom h p. We will assume that the intersection of the feasible set, C = {x : f m (x) 0, h p (x) = 0, m = 1,..., M, p = 1,..., P } is a non-empty and a subset R N. 1

2 The Lagrangian takes the constraints in the program above and integrates them into the objective function. The Lagrangian L : R N R M R P R associated with this optimization program is L(x, λ, ν) = f 0 (x) + λ m f m (x) + ν p h p (x) The x above are referred to as primal variables, and the λ, ν as either dual variables or Lagrange multipliers. The Lagrange dual function g(λ, ν) : R M R P R is the minimum of the Lagrangian over all values of x: ( ) g(λ, ν) = inf f 0 (x) + λ m f m (x) + ν p h p (x). x R N Since the dual is a pointwise infimum of a family of affine functions in λ, ν, g is concave regardless of whether or not the f m, h p are convex. The key fact about the dual function is that is it is everywhere a lower bound on the optimal value of the original program. If p is the optimal value for (1), then g(λ, ν) p, for all λ 0, ν R P. 2

3 This is (almost too) easy to see. For any feasible point x 0, λ m f m (x 0 ) + ν p h p (x 0 ) 0, and so L(x 0, λ, ν) f 0 (x 0 ), for all λ 0, ν R P, meaning g(λ, ν) = inf x R N L(x, λ, ν) L(x 0, λ, ν) f 0 (x 0 ). Since this holds for all feasible x 0, g(λ, ν) inf x C f 0 (x) = p. The (Lagrange) dual to the optimization program (1) is maximize g(λ, ν) subject to λ 0. (2) λ R M, ν R P The dual optimal value d is d = sup g(λ, ν) = sup λ 0, ν λ 0, ν inf L(x, λ, ν). x R N Since g(λ, ν) p, we know that d p. The quantity p d is called the duality gap. If p = d, then we say that (1) and (2) exhibit strong duality. 3

4 Certificates of (sub)optimality Any dual feasible 1 (λ, ν) gives us a lower bound on p, since g(λ, ν) p. If we have a primal feasible x, then we know that f 0 (x) p f 0 (x) g(λ, ν). We will refer to f 0 (x) g(λ, ν) as the duality gap for primal/dual (feasible) pair x, λ, ν. We know that p [g(λ, ν), f 0 (x)], and likewise d [g(λ, ν), f 0 (x)]. If we are ever able to reduce this gap to zero, then we know that x is primal optimal, and λ, ν are dual optimal. There are certain kinds of primal-dual algorithms that produce a series of (feasible) points x (k), λ (k), ν (k) at every iteration. We can then use f 0 (x (k) ) g(λ (k), ν (k) ) ɛ, as a stopping criteria, and know that our answer would yield an objective value no further than ɛ from optimal. Strong duality and the KKT conditions Suppose that for a convex program, the primal optimal value p an the dual optimal value d are equal p = d. 1 We simply need λ 0 for (λ, ν) to be dual feasible. 4

5 If x is a primal optimal point and λ, ν is a dual optimal point, then we must have f 0 (x ) = g(λ, ν ) ( = inf x R N f 0 (x ) + f 0 (x ). f 0 (x) + λ mf m (x) + λ mf m (x ) + ) νph p (x) νph p (x ) The last inequality follows from the fact that λ m 0 (dual feasibility), f m (x ) 0, and h p (x ) = 0 (primal feasibility). Since we started out and ended up with the same thing, all of the things above must be equal, and so λ mf m (x ) = 0, m = 1,..., M. Also, since we know x is a minimizer of L(x, λ, ν ) (second equality above), which is an unconstrained convex function (with λ, ν fixed), the gradient with respect to x must be zero: x L(x, λ, ν ) = f 0 (x )+ λ m f m (x )+ νp h p (x ) = 0. Thus strong duality immediately leads to the KKT conditions holding at the solution. Also, if you can find x, λ, ν that obey the KKT conditions, not only do you know that you have a primal optimal point on your hands, but also we have strong duality (and λ, ν are dual optimal). For if KKT holds, x L(x, λ, ν ) = 0, 5

6 meaning that x is a minimizer of L(x, λ, ν ), i.e. thus L(x, λ, ν ) L(x, λ, ν ), g(λ, ν ) = L(x, λ, ν ) = f 0 (x ) + λ mf m (x ) + = f 0 (x ), (by KKT), νph p (x ) and we have strong duality. The upshot of this is that the conditions for strong duality are essentially the same as those under which KKT is necessary. The program (1) and its dual (2) have strong duality if the f m are affine inequality constraints, or there is an x R N such that for all the f i which are not affine we have f i (x) < 0. 6

7 Examples 1. Inequality LP. Calculate the dual of minimize x, c subject to Ax b. x R N Answer: The Lagrangian is L(x, λ) = x, c + λ m ( x, a m b m ) = c T x λ T b + λ T Ax. This is a linear functional in x it is unbounded below unless c + A T λ = 0. Thus g(λ) = inf x ( ) c T x λ T b + λ T Ax = { λ, b, c + A T λ = 0, otherwise. So the Lagrange dual program is maximize λ, b subject to A T λ = c λ R M λ 0. 7

8 2. Standard form LP. Calculate the dual of minimize x R N x, c subject to Ax = b λ 0. 8

9 Least-squares. Calculate the dual of minimize x R N x 2 2 subject to Ax = b. Check that the duality gap is zero. Answer: maximize ν R M 1 4 νt AA T ν b T ν 9

10 3. Minimum norm. Calculate the dual of minimize x subject to Ax = b, x R N where is a general valid norm. Answer: Use f 0 (x) = x to ease notation below. We start with the Lagrangian: L(x, ν) = f 0 (x) + ν p ( x, a m b m ) = f 0 (x) ν, b + (A T ν) T x and so g(ν) = ν, b + inf x = ν, b sup x = ν, b f 0 ( A T ν), ( ) f 0 (x) + (A T ν) T x ( f 0 (x) (A T ν) T x ) where f 0 is the Fenchel dual of f 0 : f 0 (y) = sup( x, y f 0 (x)). x With f 0 =, we know already that so f 0 (y) = g(ν) = { 0, y 1,, otherwise, { ν, b, A T ν 1, otherwise. 10

11 Thus the dual program is maximize ν, b subject to A T ν 1, ν R P where is the dual norm of. 11